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Precalculus 11 Online
OpenStudy (anonymous):

Use the Laws of Logarithms to expand the expression.

OpenStudy (anonymous):

Where is the expression?

OpenStudy (anonymous):

\[In \sqrt{z}\]

OpenStudy (anonymous):

who can solve it??

OpenStudy (anonymous):

\(In\) you mean \(\ln\)?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[\ln \sqrt{z} = \ln (z^{1/2}) = \frac{1}{2}\ln z \]

OpenStudy (anonymous):

That simplifies the expression a bit, but I'm not completely sure of the goal here.

OpenStudy (anonymous):

correct!

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